How to test if an executable exists in the %PATH% from a windows batch file?
for %%X in (myExecutable.exe) do (set FOUND=%%~$PATH:X)
if defined FOUND ...
If you need this for different extensions, just iterate over PATHEXT
:
set FOUND=
for %%e in (%PATHEXT%) do (
for %%X in (myExecutable%%e) do (
if not defined FOUND (
set FOUND=%%~$PATH:X
)
)
)
Could be that where
also exists already on legacy Windows versions, but I don't have access to one, so I cannot tell. On my machine the following also works:
where myExecutable
and returns with a non-zero exit code if it couldn't be found. In a batch you probably also want to redirect output to NUL
, though.
Keep in mind
Parsing in batch (.bat
) files and on the command line differs (because batch files have %0
–%9
), so you have to double the %
there. On the command line this isn't necessary, so for variables are just %X
.
Windows Vista and later versions ship with a program called where.exe
that searches for programs in the path. It works like this:
D:\>where notepad
C:\Windows\System32\notepad.exe
C:\Windows\notepad.exe
D:\>where where
C:\Windows\System32\where.exe
For use in a batch file you can use the /q
switch, which just sets ERRORLEVEL
and doesn't produce any output.
where /q myapplication
IF ERRORLEVEL 1 (
ECHO The application is missing. Ensure it is installed and placed in your PATH.
EXIT /B
) ELSE (
ECHO Application exists. Let's go!
)
Or a simple (but less readable) shorthand version that prints the message and exits your app:
where /q myapplication || ECHO Cound not find app. && EXIT /B